Re: Reactionless Drives Redux

Jonathan L Cunningham wrote:
On Sat, 25 Jul 2009 18:43:53 -0700, Erik Max Francis <max@xxxxxxxxxxx>
wrote:
Jonathan L Cunningham wrote:
I think I could do the calculation (without using GR) based on their
(simple) three point mass example on the surface of a sphere... which
would presumably give an order of magnitude estimate. But what's the
approximate value of the (gravitational) radius of curvature of
space-time where we are?
You want the curvature scalar, which is (for all intents and purposes) +-1, depending on the curvature.

I thought it might be possible to get a rough estimate of the curvature
in conventional units by considering the curvature of a falling body,
i.e. it falls 9.8m in 1s, and then use c to convert both quantities to
times or both to distances.

I've thought about it a little more, it's possible to pull out conventional units. It's just a huge pain in the ass because of the conventional use of geometrized units in general relativity. (In fact, geometrized units are used precisely because keeping the units right becomes a huge pain in the ass when you're already dealing with mathematics that's not exactly two plus two). You often end up with equations that absolutely, positively look like they have the wrong units, but don't.

It's dependent on the curvature scalar (a.k.a. the Ricci scalar; same thing) R, which is zero for the Schwarzschild metric (outside of the spherically symmetric mass whose gravitational field is represented by that metric).

What surely dominates here would be the global curvature of the Universe. In a Friedmann-Robertson-Walker metric -- our standard model of a spatially homogeneous and isotropic universe -- the curvature scalar is (geometrized units):

R = (6/a^2) [d^2 a/dt^2 a + (da/dt)^2 + k],

where a(t) is the scale factor and k = +-1 depending on the curvature. (In general FRW metrics, k can also be zero, but that the assumptions of the swimming article preclude that.) Normally you only care about changes in the scale factor, so you make it dimensionless and set it equal to a(t = now = 0) = 1. (The scale factor is _not_ the current size of the observable Universe -- it's the radius of curvature of spacetime.)

The first term is from the acceleration of the Universe. The second term is from the the expansion of the Universe, and the third term is from the inherent curvature of the Universe. We can put all of these in terms of the Hubble "constant" (parameter, really) H = (da/dt)/a except the last term, which is still a function of the naked scale factor:

R = 6 (1 - q) H^2 + 6 k/a^2],

where q is the deceleration parameter, H is the Hubble "constant", and a is the scale factor. We're all interested in these values now, which are easy to come by except for a. q ~= 1/2, H = 71 km/s/Mpc, and we're not sure what a is but it's much, much bigger than the size of the observable Universe, otherwise it wouldn't look pretty flat. We can take the worst case scenario and say that it's _at least_ the size of the observable Universe, or about 14 Glyr, which will at least give us an upper estimate.

Now people paying attention will note that this is clearly dimensionally inconsistent; the first term has units of inverse square time, but the second has units of inverse square distance. That's because it's written in geometrized units, which are defined such as c = 1, and distinctions between units of length and time (and mass for that matter!) are not made; they're all measured in units of (wait for it) centimeters. (That's just an artifact of when geometrized units became popularized, most engineers were still using c.g.s. instead of SI.) So we'll figure out what units we really want and normalize things with factors of c (c^2, actually).

So back to the swimming article, the swimming triangle part of the article assumes two "arms" (with uniformly distributed weight along their length), joined together and with two weights on the the ends, forming an isosceles triangle of height h and base b. The "swimming" motion they calculate assumes that each end makes a closed motion in a little box of size db by dh. That then results in a total "swim" displacement of the triangle by a displacement magnitude dx:

dx ~= (1/2) R b dh db.

(Note they write it without the R, but that's because in geometrized units where you're assuming a(t) is dimensionless, and so R is effectively 1 and drops out.)

Note that's all still multiplied by R. Now consider what R must be in reasonable units. The geometrized version above is still dimensionally inconsistent for our purposes, but we need to know what units to convert them to. The dimensionally consistent version of this equation would require that R has units of inverse square distance (not inverse square time). So convert H^2 to units of m^-2 and find that it's a whopping 3.4 x 10^-59 m^-2. So right away we should realize how ridiculously small the effect is.

We still don't know what a is now, but let's make the outrageous assumption that it's roughly the same size of the observable Universe (it's surely much larger, so that's wrong, but think of it as a conservative estimate). Then our second term for R is (plus or minus) 5.7 x 10^-53 m^-2, much, much greater than the contribution of the first term. So in this conservative overestimate, we can safely ignore the first term. Thus we will conclude that R is (outrageous overestimate) is +- ~10^-52 m^-2 (don't forget that factor of 6). We don't really care what the sign is (though we think it should be negative since the Universe looks open), but the sign doesn't matter anyway, since we're only interested in the displacement.

So what's this mean? The displacement, once again, is

dx ~= (1/2) R b dh db.

If we have an absolutely huge triangle with a base of b = 1000 km that makes flopping motions where dh = db = 1 km, the total displacement we're going to get each complete stroke is (hold on to your butts) ~10^-40 m. Yeah. Really.

If your triangle were the size of the Milky Way Galaxy (b = 100 klyr) and your strokes were 100 lyr along each axis (roughly the distance between the Sun and, say, Algol), then the total displacement for each stroke would be ~10^-11 m, roughly the size of a hydrogen atom. Yeah. Really.

Then you can be silly and ask how long would it take. If the swimmer can move its endpoints along those dh and db distances through a full circuit at a speed of v, then the total time it would take is 2 (db + dh)/v. So the average time would be:

dx v/[2 (db + dh)].

Even if you could travel at the maximum permissible speed so that v ~= c (though one must note that if you could move things that fast, the obvious next question would be what in the world are you thinking?!), our Milky Way-sized swimmer maintains a mean swimming speed of ~10^-21 m/s or ~10^-30 c. Not so impressive are you now, Mr. Galaxy-Sized Swimming Machine?

Just because the effect exists (and it does) doesn't mean it's practical. The local spacetime curvature is fantastically low.

It would also be nice to know if the effect becomes large enough to be
useful, e.g. near a neutron star.

Nope, there would be no effect from being near a localized mass (but outside of it). The curvature scalar R (on which this effect depends) is zero outside the neutron star (or any mass distribution), because the stress-energy tensor is zero there. (Well, it wouldn't be zero if there were, say, gravitational radiation being emitted by the object, which is probably likely, but it'd be even ridiculously lower than the things we're talking about here.) Weird, huh?

--
Erik Max Francis && max@xxxxxxxxxxx && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM/Y!M/Skype erikmaxfrancis
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